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Math Help - should be simple

  1. #1
    Junior Member winterwyrm's Avatar
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    should be simple

    A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 42 ft/s2. What is the distance covered before the car comes to a stop? (Give your answers correct to one decimal place.)

    I think the unit conversion is what keeps making my answer wrong....

    Oh, and thanks in advance.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by winterwyrm View Post
    A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 42 ft/s2. What is the distance covered before the car comes to a stop? (Give your answers correct to one decimal place.)

    I think the unit conversion is what keeps making my answer wrong....

    Oh, and thanks in advance.
    Let's find out if it is the units.

    Set up a coordinate system where the origin is where the car starts to brake and a +x direction in the direction of the velocity of the car at the origin.

    This is what we know:
    t = ?
    x_0 = 0~ft x = ?
    v_0 = 50~mi/h v = 0~ft/s
    a = -42~ft/s^2

    So let's first convert that initial speed:
    \frac{50~mi}{1~h} \cdot \frac{5280~ft}{1~mi} \cdot \frac{1~h}{3600~s} = 73.33~ft/s

    We want x. So
    v^2 = v_0^2 + 2a(x - x_0)

    0^2 = v_0^2 + 2ax

    x = -\frac{v_0^2}{2a}

    I get x = 64.0212 ft.

    -Dan
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  3. #3
    Junior Member winterwyrm's Avatar
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    yup it was the units, I was trying to convert the other way around, and just threw a ratio in there, apparently in the wrong place.... Thanks a ton!

    One thing though, where did you get that equation?
    Last edited by winterwyrm; November 17th 2007 at 01:42 PM.
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  4. #4
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    Hello, winterwyrm!

    Another approach . . .


    A car is traveling at 50 mi/hr when the brakes are fully applied,
    producing a constant deceleration of 42 ft/sē.
    What is the distance covered before the car comes to a stop?
    (Give your answers correct to one decimal place.)

    Acceleration is the derivative of velocity.

    We have: . a(t) \:=\:v'(t) \:=\:-42

    Integrate: . v(t) \:=\:-42t +C_1

    When t = 0,\;v(0) \:=\:50\text{ mi/hr} \:=\:\frac{220}{3}\text{ ft/sec}
    . . So we have: . \frac{220}{3} \:=\:-42(0) + C_1\quad\Rightarrow\quad C_1 \:=\:\frac{220}{3}
    . . Hence: . v(t) \:=\:-42t + \frac{220}{3} . [1]


    Integrate: . s(t) \:=\:-21t^2 + \frac{220}{3}t + C_2

    When t = 0,\;s(0) \:=\:0
    . . So we have: . 0 \:=\:-21(0^2) + \frac{220}{3}(0) + C_2\quad\Rightarrow\quad C_2 \:=\:0
    . . Hence: . s(t) \;=\;-21t^2 + \frac{220}{3}t . [2]


    The car comes to a stop when v(t) = 0.
    . . From [1]: . -42t + \frac{220}{3} \;=\;0\quad\Rightarrow\quad t \;=\;\frac{110}{63} seconds.

    Substitute into [2]: . s\!\left(\frac{110}{63}\right) \;=\;-21\left(\frac{110}{63}\right)^2 + \frac{220}{3}\left(\frac{110}{63}\right) \;=\;\frac{12,100}{189} \;\approx\; 64.0212\text{ ft}

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